derivative of 1/(z^2+1), chain rule
derivative of 1/(z^2+1), chain rule

Example 27 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

Examples

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15 Important

Example 16

Example 17 Important

Example 18

Example 19

Example 20 Important

Example 21

Example 22

Example 23

Example 24 Important

Example 25

Example 26 (i)

Example 26 (ii) Important

Example 26 (iii) Important

Example 26 (iv)

Example 27 Important

Example 28

Example 29 Important

Example 30 Important

Example 31

Example 32

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37

Example 38 Important

Example 39 (i)

Example 39 (ii) Important

Example 40 (i)

Example 40 (ii) Important

Example 40 (iii) Important

Example 41

Example 42 Important

Example 43

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams You are here

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Question 3 Find the derivative of f given by f (x) = tan–1 𝑥 assuming it exists. 𝑓 (𝑥)=〖𝑡𝑎𝑛〗^(−1) 𝑥 Let 𝒚= 〖𝒕𝒂𝒏〗^(−𝟏) 𝒙 tan〖𝑦=𝑥〗 𝒙=𝐭𝐚𝐧〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (tan𝑦 ))/𝑑𝑥 1 = (𝑑 (tan𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (tan𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = 〖𝐬𝐞𝐜〗^𝟐 𝒚 . 𝑑𝑦/𝑑𝑥 1 = (𝟏 + 𝒕𝒂𝒏𝟐𝒚) 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(1 + 〖𝐭𝐚𝐧〗^𝟐𝒚 ) Putting 𝑡𝑎𝑛𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = 1/(1 + 𝒙^𝟐 ) Hence (𝒅(〖𝐭𝐚𝐧〗^(−𝟏)〖𝒙)〗)/𝒅𝒙 = 𝟏/(𝟏 + 𝒙^𝟐 ) As 𝑦 = 〖𝑡𝑎𝑛〗^(−1) 𝑥 So, 𝒕𝒂𝒏𝒚 = 𝒙 Derivative of 〖𝒄𝒐𝒔〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑠〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒔〗^(−𝟏) 𝒙 cos〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐬〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cos𝑦 ))/𝑑𝑥 1 = (𝑑 (cos𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cos𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = (−sin𝑦) 𝑑𝑦/𝑑𝑥 (−1)/sin𝑦 =𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (−1)/𝒔𝒊𝒏𝒚 𝑑𝑦/𝑑𝑥= (−1)/√(𝟏 − 〖𝒄𝒐𝒔〗^𝟐 𝒚) Putting 𝑐𝑜𝑠〖𝑦=𝑥〗 𝑑𝑦/𝑑𝑥= (−1)/√(1 − 𝒙^𝟐 ) Hence, (𝒅(〖𝒄𝒐𝒔〗^(−𝟏) 𝒙” ” ))/𝒅𝒙 = (−𝟏)/√(𝟏 − 𝒙^𝟐 ) “We know that” 〖𝑠𝑖𝑛〗^2 𝜃+〖𝑐𝑜𝑠〗^2 𝜃=1 〖𝑠𝑖𝑛〗^2 𝜃=1−〖𝑐𝑜𝑠〗^2 𝜃 𝒔𝒊𝒏𝜽=√(𝟏−〖𝒄𝒐𝒔〗^𝟐 𝜽) ” ” As 𝑦 = 〖𝑐𝑜𝑠〗^(−1) 𝑥 So, 𝒄𝒐𝒔𝒚 = 𝒙 Derivative of 〖𝒄𝒐𝒕〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑡〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒕〗^(−𝟏) 𝒙 cot〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐭〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cot𝑦 ))/𝑑𝑥 1 = (𝑑 (cot𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (cot𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = −𝐜𝐨〖𝐬𝐞𝐜〗^𝟐 𝒚 . 𝑑𝑦/𝑑𝑥 1 = −(𝟏 +𝒄𝒐𝒕𝟐𝒚) 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(1 + 〖𝐜𝐨𝐭〗^𝟐𝒚 ) Putting 𝑐𝑜𝑡𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(𝒙^𝟐 + 𝟏) Hence (𝒅(〖𝐜𝐨𝐭〗^(−𝟏)〖𝒙)〗)/𝒅𝒙 = (−𝟏)/(𝒙^𝟐 + 𝟏) (𝐴𝑠 〖 𝑐𝑜𝑠𝑒𝑐〗^2〖𝑦= 〖1+〗〖𝑐𝑜𝑡〗^2𝑦 〗) As 𝑦 = 〖𝑐𝑜𝑡〗^(−1) 𝑥 So, 𝒄𝒐𝒕𝒚 = 𝒙 Derivative of 〖𝒔𝒆𝒄〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑠𝑒𝑐〗^(−1) 𝑥 Let 𝒚= 〖𝒔𝒆𝒄〗^(−𝟏) 𝒙 sec〖𝑦=𝑥〗 𝒙=𝐬𝐞𝐜〖𝒚 〗 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (sec𝑦 ))/𝑑𝑥 1 = (𝑑 (sec𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 1 = (𝑑 (sec𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = 𝒕𝒂𝒏𝒚 .𝒔𝒆𝒄𝒚. 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(𝒕𝒂𝒏𝒚 .〖 sec〗𝑦 ) 𝑑𝑦/𝑑𝑥 = 1/((√(〖𝐬𝐞𝐜〗^𝟐𝒚 − 𝟏)) .〖 sec〗𝑦 ) Putting value of 𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = 1/((√(𝑥^2 − 1 ) ) . 𝑥) 𝑑𝑦/𝑑𝑥 = 1/(𝑥 √(𝑥^2 − 1 ) ) Hence 𝒅(〖𝒔𝒆𝒄〗^(–𝟏) 𝒙)/𝒅𝒙 = 𝟏/(𝒙 √(𝒙^𝟐 − 𝟏 ) ) As tan2 θ = sec2 θ – 1, tan θ = √(“sec2 θ – 1” ) As 𝑦 = 〖𝑠𝑒𝑐〗^(−1) 𝑥 So, 𝒔𝒆𝒄𝒚 = 𝒙Derivative of 〖𝒄𝒐𝒔𝒆𝒄〗^(−𝟏) 𝒙 𝑓 (𝑥)=〖𝑐𝑜𝑠𝑒𝑐〗^(−1) 𝑥 Let 𝒚= 〖𝒄𝒐𝒔𝒆𝒄〗^(−𝟏) 𝒙 cosec〖𝑦=𝑥〗 𝒙=𝐜𝐨𝐬𝐞𝐜〖𝒚 〗 1 = (𝑑 (cosec𝑦 ))/𝑑𝑦 × 𝑑𝑦/𝑑𝑥 1 = −cosec𝑦 .cot𝑦 . 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 1/(〖−cosec〗𝑦 .𝒄𝒐𝒕𝒚 ) 𝑑𝑦/𝑑𝑥 = 1/(〖−cosec〗𝑦 . √(〖𝐜𝐨𝒔𝒆𝒄〗^𝟐𝒚 − 𝟏)) Putting value of 𝑐𝑜𝑠𝑒𝑐𝑦 = 𝑥 𝑑𝑦/𝑑𝑥 = (−1)/(𝑥 √(𝑥^2 − 1 ) ) Hence 𝒅(〖𝒄𝒐𝒔𝒆𝒄〗^(–𝟏) 𝒙)/𝒅𝒙 = (−𝟏)/(𝒙 √(𝒙^𝟐 − 𝟏 ) ) As cot2 θ = cosec2 θ – 1, cot θ = √(“cosec2 θ – 1” ) Differentiating both sides 𝑤.𝑟.𝑡.𝑥 𝑑𝑥/𝑑𝑥 = (𝑑 (cosec𝑦 ))/𝑑𝑥 1 = (𝑑 (cosec𝑦 ))/𝑑𝑥 × 𝑑𝑦/𝑑𝑦 As cot2 θ = cosec2 θ – 1, cot θ = √(“cosec2 θ – 1” ) As 𝑦 = co〖𝑠𝑒𝑐〗^(−1) 𝑥 So, co𝒔𝒆𝒄𝒚 = 𝒙

You are watching: Find derivative of f(x) = tan-1 x. Info created by Bút Chì Xanh selection and synthesis along with other related topics.